You have found the following ages (in years) of all 6 zebras at your local zoo: $ 8,\enspace 5,\enspace 2,\enspace 6,\enspace 6,\enspace 15$ What is the average age of the zebras at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{8 + 5 + 2 + 6 + 6 + 15}{{6}} = {7\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $8$ years $1$ year $1$ year $^2$ $5$ years $-2$ years $4$ years $^2$ $2$ years $-5$ years $25$ years $^2$ $6$ years $-1$ years $1$ year $^2$ $6$ years $-1$ years $1$ year $^2$ $15$ years $8$ years $64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{1} + {4} + {25} + {1} + {1} + {64}} {{6}} $ $ {\sigma^2} = \dfrac{{96}}{{6}} = {16\text{ years}^2} $ The average zebra at the zoo is 7 years old. The population variance is 16 years $^2$.